# Deriviative of a log

#### vipertongn

1. The problem statement, all variables and given/known data

find dy/dx of log3*arccsc2^x

3. The attempt at a solution

I got up to 1/ln3csc2^x * 1/2^x(sqrt(2^2x-1)

However there's also a 2^x*ln2 after that

How is that?

In otherwords how would I derive 2^x would i treat it like e^x or something?

Also if its x^x why would that be 1+lnx?

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#### mistermath

I don't feel like reading your main question; it just looks messy. However, I can answer your sub q I think!

d/dx of e^x is e^x * ln (e) = e^x and d/dx of 2^x is 2^x ln (2).

To derive this, suppose y = 2^x, then ln y = ln (2^x) = x ln (2).

derivative of both sides gives you:
1/y * dy/dx = ln (2) multiply both sides by y to get:

dy/dx = y ln (2) and substitute y = 2^x to get

dy/dx = 2^x ln (2)

I also did x ^ x the same way, try it before looking at my solution..

let y = x^x then
ln y = x ln (x)

1/y * dy/dx = 1 * ln (x) + 1/x * x = ln (x) + 1 ->

dy/dx = y ln(x) + y = x^x * ln (x) + x^x or more commonly x^x * [ln (x) + 1]

#### vipertongn

Thanks!! that's all i needed

#### HallsofIvy

1. The problem statement, all variables and given/known data

find dy/dx of log3*arccsc2^x
That is log3(arcsec(2x))?

3. The attempt at a solution

I got up to 1/ln3csc2^x * 1/2^x(sqrt(2^2x-1)
I have no idea how you got that! How did "arcsec" become "csc"? The derivative of log x is 1/x so you should have 1/(arcsec(2x)) times the derivative of that denominator.

However there's also a 2^x*ln2 after that

How is that?

In otherwords how would I derive 2^x would i treat it like e^x or something?

Also if its x^x why would that be 1+lnx?
$$2^x= e^{ln(2^x)}= e^{x ln(2)}$$ so the derivative of 2x is ln(2)2x.

To differentiate y= xx, write it as ln(y)= x ln(x). On the left, the derivative is (1/y) y'. On the right, it is ln(x)+ x(1/x)= ln(x)+ 1. So y'= (ln(x)+ 1)y= (ln(x)+ 1)xx.

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